[Ford Prefect]

I have a question regarding the pressure term in the radial direction (r-momentum equation).


Why do we have dp/dr and not 1/r*d(r*p)dr ?

[FMDenaro]

Have you writted Grad p in cylindrical coordinate?


Grad = ir d/dr +itheta (1/r) d/dtheta + iz d/dz

[Ford Prefect]

Then how about the stress terms, in particular Tau_xx, which is derived on the same basis as pressure in r-direction (both acting on the same surface)?

[FMDenaro]

https://vhub.org/resources/4210/down...ylindrical.pdf

[Ford Prefect]

Thank you! This makes it a bit clearer.



I guess the "extra" term that I missed is the (sigma_theta_theta / r) in equation 11.


What is the origin of this term? Is it because the surfaces are not orthogonal, which means that there is a contribution to the force in the r-direction from the stress in the theta direction? It would seem that way by looking at equation 10.



Also from a pedagogical perspective I find it a bit confusing when mixing the Taylor expansion approach (figure 1) with the methodology described in equations 10 and 11, i.e. the figure does not convey what is done in the derivation of the equations.

[FMDenaro]

Again you can develop the expression for Div.T using the previous gradient expression and the diadic expression for the tensor. You need to apply the derivative rule to the unit vector ir, itheta and iz.

[LuckyTran]

Originally Posted by Ford Prefect

What is the origin of this term? Is it because the surfaces are not orthogonal, which means that there is a contribution to the force in the r-direction from the stress in the theta direction? It would seem that way by looking at equation 10.

Also from a pedagogical perspective I find it a bit confusing when mixing the Taylor expansion approach (figure 1) with the methodology described in equations 10 and 11, i.e. the figure does not convey what is done in the derivation of the equations.

Right. The surface at theta+dtheta is not orthogonal to the r-direction but has a projection along r. The figure does not even label this stress and maybe even mislabels it so I agree it is a bit confusing. So writing down equation 10 is confusing (if you try to use this picture here) but how to get to equation 11 from equation 10 is straightforward. I would like to complain but I know it is pain in some parts of your body to make these figures. But I don't know what's missing from a pedagogical perspective? You define a control volume and coordinate system. You do Taylor series and then sum the forces in each direction.

[Ford Prefect]

Originally Posted by LuckyTran

Right. But I don't know what's missing from a pedagogical perspective? You define a control volume and coordinate system. You do Taylor series and then sum the forces in each direction.

Yeah, but this is not what is being done in equations 10 to 11. The approach that is given in the equations use the formal definition of the derivative rather than a Taylor series expansion. Sure, the end result is the same (if we skip the higher order terms from the Taylor series expansion). So, the figure indicates something that is not used in the text and if the reader does not know about the two approaches then I think it could be very confusing. Never assume that the reader knows as much as you do


I have full respect for the often tedious work of producing high quality lecture notes, so I do not want to sound overly critical here. The notes definitely helped.

[LuckyTran]

The main steps are:

1. You apply formal taylor series to draw Figure 1 and label the forces/stresses on all the faces. The figure is missing some labels and doesn't write out the infinite series. Okay... But let's assume they did draw the figure correctly writing out the infinite taylor series.
2. You sum along each direction x,y, & z to write down Equation 10. If you draw correctly figure 1, there is no magic here.
3. Then you divide by the cell volume and take the limit to get to Equation 11.

Steps 2 & 3 (i.e. writing down eq 10 and 11) I think is clearly explained. What's missing is they don't state taylor series are used to generate Figure 1. I don't know if that's what you think is missing but it happens before eq 10 & 11.

[Ford Prefect]

You do not have to apply the Taylor series expansion do describe the value of (r+dr). You can just as well write sigma*Area(r+dr) - sigma*Area(r), etc. Then dividing by the volume and letting the sizes go to zero yields the pde. Conversely, when applying the Taylor series expansion "method" then you do not need to take the limit as the size of the control volume shrinks, you will still end up with the pde. The definition of the (first) derivative is "built in" into the Taylor series and as such it feels weird to apply it when we have expressed the (r+dr) side as a first order Taylor expansion about the r side (even though the terms cancel and we get the same end result).

[LuckyTran]

Originally Posted by Ford Prefect

You do not have to apply the Taylor series expansion do describe the value of (r+dr).

I agree you do not have to apply Taylor approximations to describe the value at r+dr, you are just writing down sigma at the r+dr location. But you simply cannot just write down sigma*Area*(r+dr). Because you know the value of sigma and the area can change! This value can be anything. It is not until you bring in Taylor series that you can say the sigma at the r+dr. You need not use Taylor approximations, you can use other methods to write down what sigma at the r+dr location is. But those would be different methods.

Originally Posted by Ford Prefect

Conversely, when applying the Taylor series expansion "method" then you do not need to take the limit as the size of the control volume shrinks, you will still end up with the pde. The definition of the (first) derivative is "built in" into the Taylor series and as such it feels weird to apply it when we have expressed the (r+dr) side as a first order Taylor expansion about the r side (even though the terms cancel and we get the same end result).

Maybe it's a matter of formalism. But the point is to start with an arbitrarily control volume and let it shrink. Taylor approximations (in discrete form) are even more general than the taylor series (in terms of derivatives). If your goal is only find a way to write down the final N-S pde's, then you might not appreciate it.

Also you are not supposed to limit yourself to a first order expansion. Taylor series means you write down the infinite series. Most people don't write beyond the 1st derivative (heck this figure even skips some of the 1st derivatives) and that's why it appears you are doing a first order expansion. It is very important to do the full series and prove that the terms cancel out, otherwise and residual error terms must be analyzed properly! Of course you already know these terms will cancel because they don't show up in the final p.d.e.

The limit definition of the derivative looks built-in for Taylor series if you are used to seeing Taylor series from calculus. You have to remember taylor approximations in discrete form predates that and is much more general. I.e. the dr in the figure should not be the differential (infinitesimal) dr but the discrete (big del) dr.

[Ford Prefect]

Perhaps we are talking about the same things, but this is how I see the difference in the methods (both of course give the same answer). The example in the link above give some sort of hybrid between these two methods (also arriving at the same answer).


Regarding how the terms in the Taylor expansion cancel out if we keep higher order terms, I am not so sure. Normally we just assume that the control volume is small enough to justify dropping the higher order terms. Apart from numerical approximation of derivatives, where terms often cancel out, could you show how you mean in this example?

[LuckyTran]

Okay I understand better now what you mean.

If you correctly label the flux at x+dx as phi(x+dx) then yes you end up with the same result. So in this case you would remake the original figure and only label the fluxes without applying taylor approximations.

But it only looks nice when you do it in cartesian because your area is not changing. In cylindrical and spherical you also get an area change and here it is not so obvious how to reduce it.


p.s. I got ahead of myself when I said the high order terms in taylor series cancel out. This is very dependent on the problem you are working with and depends where you center your coordinate system. In this example they disappear when you take the limit. If you don't take the limit and keep it discrete, the terms remain. If you don't take the limit you of course don't get any closer to the continuous navier-stokes, so it's a different exercise.

[FMDenaro]

Be only careful to the strong assumption when you switch from the difference of the fluxes to the derivative. This is the counterpart of assuming valid Gauss. Of course, that is not true if you have a singularity in the volume as happens for shock.

[Ford Prefect]

Originally Posted by LuckyTran

Okay I understand better now what you mean.

But it only looks nice when you do it in cartesian because your area is not changing. In cylindrical and spherical you also get an area change and here it is not so obvious how to reduce it.

This is how I would do it without Taylor series expansion

[LuckyTran]

Originally Posted by Ford Prefect

This is how I would do it without Taylor series expansion

Right... Notice how in the very last line you get something that is not simply d(phi)/dr but you get the divergence (as you should). I.e. you can't simply write down the navier-stokes equations from the definition of the derivative.

You also cheated in the 2nd to last line. You dropped the indices labeling the flux at r + dr when you took the limit as dr=> 0. This is valid only if phi is constant over the cell. Otherwise, you have no idea what the flux at r + dr is (yet). This doesn't show up in cartesian because you conveniently recovered the derivative (by luck & not by principle).

[Ford Prefect]

Thank you for your input. I have a few follow-up questions in which I would be happy if you could show an example or two as explanation.

Originally Posted by LuckyTran

Right... Notice how in the very last line you get something that is not simply d(phi)/dr but you get the divergence (as you should). I.e. you can't simply write down the navier-stokes equations from the definition of the derivative.

Not sure what you mean. The definition of the derivative is there, just as the additional term is. It is all part of taking the limit as the sides of the volume goes to zero. Is your point here that it is impossible to ONLY use the definition of the derivative (say if I would have skipped the extra term)?

Originally Posted by LuckyTran

You also cheated in the 2nd to last line. You dropped the indices labeling the flux at r + dr when you took the limit as dr=> 0. This is valid only if phi is constant over the cell. Otherwise, you have no idea what the flux at r + dr is (yet). This doesn't show up in cartesian because you conveniently recovered the derivative (by luck & not by principle).

Ok, I didn't know that was considered cheating. Do you mean that phi(r+delta_r) does not approach phi(r) when delta_r approaches zero?

If my example was successful, due to luck, then could you please show a counter-example instead where the above method does not work?

[FMDenaro]

As general comment, I think that any writing of the equations in a specific reference system starts from the most general formulation. Therefore, you can consider the integral form and write the volume and surface integrals according to the decomposition over normal domains.
Similarily, you can use the general vector differential formulation (provided that Gauss applies) and project the terms along the directions.


This way, any doubt about the terms is resolved.

[Ford Prefect]

Sure, I agree. I think the approach that I have shown is using this method though. I have not written out the integrals, but they can be be thought of as part of the surface area formulation and volume. I understand that the integral formulation is also a very good approach for the FV method since we can use different types of numerical integrals to approximate the fluxes with varying order of accuracy (although 2 is usually the highest used).


If we are only considering how to derive the derivative form of the equations, could you then please comment on the methodology I presented. Is it wrong?

[FMDenaro]

Originally Posted by Ford Prefect

Sure, I agree. I think the approach that I have shown is using this method though. I have not written out the integrals, but they can be be thought of as part of the surface area formulation and volume. I understand that the integral formulation is also a very good approach for the FV method since we can use different types of numerical integrals to approximate the fluxes with varying order of accuracy (although 2 is usually the highest used).


If we are only considering how to derive the derivative form of the equations, could you then please comment on the methodology I presented. Is it wrong?

Using the integral form you have also the relevant advantage that in a cylindrical system you will not have to face the singularity at r=0.
Consider the integral Int[S] n.f dS it will be decomposed in two surface integrals over the top and bottom circunferences and in a sum of the surface integrals over the most internal lateral faces around r=0.



Considering the differential form you just have to use the expression of nabla and vectors in the cilyndrical system. For example


Div v =(ir d/dr +itheta (1/r)d/dtheta + iz d/dz ).(ir vr +itheta vtheta + iz vz )


Then apply the rule of derivative of the unit vectors and perform the products